Concept:The limiting reactant must be calculated by finding the available moles of each reactant and dividing by their respective stoichiometric coefficients. The lowest resulting number limits the reaction.
Formula:$$ \text{Limiting Factor} = \frac{\text{Moles supplied}}{\text{Stoichiometric Coefficient}} $$
Solution:- Calculate Moles of Carbon Disulfide (\( \text{CS}_2 \)): Molar mass = \( 12 + (32 \times 2) = 76 \text{ g/mol} \).
- Moles of \( \text{CS}_2 = 7.6 / 76 = 0.1 \text{ moles} \).
- Calculate Moles of Oxygen (\( \text{O}_2 \)): Molar mass = \( 32 \text{ g/mol} \).
- Moles of \( \text{O}_2 = 12.8 / 32 = 0.4 \text{ moles} \).
- Divide by coefficients from the balanced equation (\( \text{CS}_2 + 3\text{O}_2 \)):
- For \( \text{CS}_2 \): \( 0.1 / 1 = 0.1 \). (Smallest value, so this limits).
- For \( \text{O}_2 \): \( 0.4 / 3 = 0.133 \). (Larger value, meaning it is in excess).
- Therefore, \( \text{CS}_2 \) is entirely consumed first and acts as the limiting reactant.
Why other options are incorrect:- \( \text{O}_2 \): Math proves it is the excess reactant.
- \( \text{CO}_2 \) & \( \text{SO}_2 \): These are the products formed by the reaction, not the reactants limiting it.
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